# Legendre's Condition

## Theorem

Let $y =\map y x$ be a real function, such that:

$\map y a = A,\quad \map y b = B$

Let $J \sqbrk y$ be a functional, such that:

$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$

where

$F \in C^2 \closedint a b$

with respect to all its variables, and $C$ stands for differentiability class.

Then a necessary condition for $J \sqbrk y$ to have a minimum at $y = \hat y$ is:

$\bigintlimits {F_{y'y'} } {y \mathop = \hat y} {} \ge 0 \quad \forall x \in \closedint a b$

## Proof

### Lemma 1

Let $y = \map y x$ be a real function, such that:

$\map y a = A, \quad \map y b = B$

Let $J \sqbrk y$ be a functional, such that:

$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$

where:

$F \in C^2 \closedint a b$

with respect to all its variables.

Then:

$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x } h'^2 + \map Q {x, \map y x} h^2} \rd x$

where:

$\map P {x, \map y x} = \dfrac 1 2 F_{y' y'}, \quad \map Q {x, \map y x} = \dfrac 1 2 \paren {F_{yy} - \dfrac \d {\d x} F_{yy'} }$

### Lemma 2

Let $h$ be a real function such that:

$h \in C^1 \openint a b, \quad \map h a = 0, \quad \map h b = 0$

Let:

$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x} h'^2 + \map Q {x, \map y x} h^2} \rd x$

where $P \in C^0 \closedint a b$.

Then a necessary condition for:

$\delta^2 J \sqbrk {y; h} \ge 0$

is:

$\map P {x, \map y x} \ge 0 \quad \forall x \in \closedint a b$

$\blacksquare$

## Source of Name

This entry was named for Adrien-Marie Legendre.